Patrick Stevens

Recently, a friend re-introduced me to the joys of the nonogram (variously known as “hanjie” or “griddler”). I was first shown these about ten years ago, I think, because they appeared in The Times. When The Times stopped printing them, I forgot about them for a long time, until two years ago, or thereabouts, I tried these on a website. I find the process much more satisfying on paper with a pencil than on computer, so I gave them up again and forgot about them again.

Anyway, the thought occurred to me: is a given griddler always solvable, and is it solvable uniquely? That is, given a grid and the edge entries, is it always a valid puzzle?

Notation: we will say that a given solved grid has an edge-set consisting of the numbers we would see if we were about to start solving the nonogram. We say that an edge-set applies to a solved grid if that edge-set is consistent with the solved grid. (For instance, the empty edge-set doesn’t apply to any solved grid apart from the zero-size grid.)

Then our question has become: is there in some way a bijection between (edge-sets) and (solved grids)?

Existence of edge-sets

We can trivially describe any solved grid by an edge-set and a grid size: simply write down the grid size of the solved grid, and write down the obvious edge-set. (We do need the grid size to be specified, because given an edge-set which applies to a solved grid, we can create a new grid to which that edge-set applies by simply appending a blank row to the solved grid.)

Uniqueness of edge-sets

Is there an obvious reason why we could never have two different edge-sets applying to the same solved grid? It seems intuitively clear that a given solved grid can only have the obvious edge-set (namely, the one we get by writing down the blocks in each row and column in the obvious way). Is this rigorous as a proof? Yes: suppose that we had two edge-sets describing the same solved grid, and (wlog) the sets differ in the first row. In fact, let us wlog that our solved grid is only one row long.

If one edge-set is empty, we’re done: because the two edge-sets are not the same, that means the other edge-set is non-empty, and so under the first edge-set the solved grid is empty, while under the second the solved grid is nonempty.

If both edge-sets are non-empty: suppose the first starts with the number , and the second with the number . Then we have some number of blank squares, and then filled-in squares (by edge-set 1) and also filled-in squares (by edge-set 2); hence , because our solved grid is fixed.

Existence of solutions

Must a solution exist for a given grid size and edge-set? Is it possible to create a nonogram with no solution? One strategy for proving this might be to count the number of allowable edge-sets and to count the number of allowable solved grids (the latter problem is extremely easy if we consider a grid as being a binary number whose bits are laid out in a rectangle), because we have that any two finite sets of the same size must biject. However, the former problem sounds very hard.

On second thoughts (read: I slept on this), it’s blindingly obvious that there is a grid with no solution - namely, the one-by-one grid with edge-set “1 as column heading, 0 as row heading”. So there certainly are edge-sets which don’t have a solution grid.

Uniqueness of solutions

OK, if we don’t always have solvability, how about the “easy puzzle-setting property”: that a given edge-set and grid-size cannot have two solved grids to which the edge-set applies? If this were true, it would make generating puzzles extremely easy: simply draw out a solved grid, write down its edge-set (which is unique, as shown above), and set that edge-set and grid-size as the puzzle, without fear that someone could sit down and solve the puzzle validly to get a different grid to your solution.

On the same second thoughts as the ‘existence of solutions’ thoughts, it’s clear that the 2-by-2 grid with a diagonal black stripe has two solutions - namely, send the stripe top-left to bottom-right, or top-right to bottom-left. Curses.

Summary

Every solved grid has an edge-set, which is unique to that grid. However, not all edge-sets are solvable, and we don’t have uniqueness of solutions. That was much less interesting than I had hoped.